PlanetMath: basis of ideal in algebraic number field

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basis of ideal in algebraic number field

(Theorem)

Theorem. Let $\mathcal{O}_K$ be the maximal order of the algebraic number field of degree . Every ideal $\mathfrak{a}$ of $\mathcal{O}_K$ has a basis, i.e. there are in $\mathfrak{a}$ the linearly independent numbers $\alpha_1,\,\alpha_2,\,\ldots,\,\alpha_n$ such that the numbers

$\displaystyle m_1\alpha_1+m_2\alpha_2+\ldots+m_n\alpha_n,$

where the

's run all rational integers, form precisely all numbers of $\mathfrak{a}$ . One has also

$\displaystyle \mathfrak{a} = (\alpha_1,\,\alpha_2,\,\ldots,\,\alpha_n),$

i.e. the basis of the ideal can be taken for the system of generators of the ideal.

Since $\{\alpha_1,\,\alpha_2,\,\ldots,\,\alpha_n\}$ is a basis of the field extension $K/\mathbb{Q}$ , any element of $\mathfrak{a}$ is uniquely expressible in the form $m_1\alpha_1+m_2\alpha_2+\ldots+m_n\alpha_n$ .

It may be proven that all bases of an ideal $\mathfrak{a}$ have the same discriminant $\Delta(\alpha_1,\,\alpha_2,\,\ldots,\,\alpha_n)$ , which is an integer; it is called the discriminant of the ideal. The discriminant of the ideal has the minimality property, that if $\beta_1,\,\beta_2,\,\ldots,\,\beta_n$ are some elements of $\mathfrak{a}$ , then

$\displaystyle \Delta(\beta_1,\,\beta_2,\,\ldots,\,\beta_n) \geqq \Delta(\alpha_1,\,\alpha_2,\,\ldots,\,\alpha_n)$ or $\displaystyle \quad \Delta(\beta_1,\,\beta_2,\,\ldots,\,\beta_n) = 0$

But if $\Delta(\beta_1,\,\beta_2,\,\ldots,\,\beta_n) = \Delta(\alpha_1,\,\alpha_2,\,\ldots,\,\alpha_n)$ , then also the $\beta_i$ 's form a basis of the ideal $\mathfrak{a}$ .

Example. The integers of the quadratic field $\mathbb{Q}(\sqrt{2})$ are $l+m\sqrt{2}$ with $l,\,m \in \mathbb{Z}$ . Determine a basis $\{\alpha_1,\,\alpha_2\}$ and the discriminant of the ideal a) $(6\!-\!6\sqrt{2},\,9\!+\!6\sqrt{2})$ , b) $(1\!-\!2\sqrt{2})$ .

a) The ideal may be seen to be the principal ideal , since the both generators are of the form $(l+m\sqrt{2})\cdot3$ and on the other side, $3 = 0\cdot(6-6\sqrt{2})+(3-2\sqrt{2})(9+6\sqrt{2})$ . Accordingly, any element of the ideal are of the form

$\displaystyle (m_1+m_2\sqrt{2})\cdot3 = m_1\cdot3+m_2\cdot3\sqrt{2}$

where

and

are rational integers. Thus we can infer that

$\displaystyle \alpha_1 = 3, \quad \alpha_2 = 3\sqrt{2}$

is a basis of the ideal concerned. So its discriminant is

$\displaystyle \Delta(\alpha_1,\,\alpha_2) = \left\vert\begin{matrix} 3 & 3\sqrt{2}\ 3 & -3\sqrt{2} \end{matrix}\right\vert^2 = 648.$

b) All elements of the ideal $(1-2\sqrt{2})$ have the form

$\displaystyle \alpha = (a+b\sqrt{2})(1-2\sqrt{2}) = (a-4b)+(b-2a)\sqrt{2}$ with $\displaystyle a,\,b \in\mathbb{Z}.$

(1)

Especially the rational integers of the ideal satisfy

, when

and thus $\alpha = a-4\cdot2a = -7a$ . This means that in the presentation $\alpha = m_1\alpha_1+m_2\alpha_2$ we can assume $\alpha_1$ to be

. Now the rational portion

in the form (1) of $\alpha$ should be splitted into two parts so that the first would be always divisible by 7 and the second by

, i.e.

; this equation may be written also as

$\displaystyle (2x+1)a-(x+4)b = 7m_1.$

By experimenting, one finds the simplest value

, another would be

. The first of these yields

$\displaystyle \alpha = 7(a-b)+(b-2a)(3+\sqrt{2}) = m_1\cdot7+m_2(3+\sqrt{2}),$

i.e. we have the basis

$\displaystyle \alpha_1 = 7, \quad \alpha_2 = 3+\sqrt{2}.$

The second alternative

similarly would give

$\displaystyle \alpha_1 = 7, \quad \alpha_2 = 10+\sqrt{2}.$

For both alternatives, $\Delta(\alpha_1,\,\alpha_2) = 392$ .

"basis of ideal in algebraic number field" is owned by pahio.

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Other names:

basis of ideal in number field

Also defines:

basis of ideal, ideal basis, discriminant of the ideal

This object's parent.

Attachments:: proof of basis of ideal in algebraic number field (Proof) by lalberti

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Cross-references: equation, divisible, presentation, side, principal ideal, quadratic field, property, discriminant, bases, expressible, field extension, generators, integers, rational, numbers, linearly independent, basis, ideal, degree, algebraic number field, maximal order
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This is version 6 of basis of ideal in algebraic number field, born on 2008-02-24, modified 2008-03-26.
Object id is 10329, canonical name is BasisOfIdealInAlgebraicNumberField.
Accessed 658 times total.

Classification:

AMS MSC:	06B10 (Order, lattices, ordered algebraic structures :: Lattices :: Ideals, congruence relations)
	11R04 (Number theory :: Algebraic number theory: global fields :: Algebraic numbers; rings of algebraic integers)
	12F05 (Field theory and polynomials :: Field extensions :: Algebraic extensions)

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